'''
There are 2N people a company is planning to interview. The cost of flying the i-th person to city A is costs[i][0], and the cost of flying the i-th person to city B is costs[i][1].

Return the minimum cost to fly every person to a city such that exactly N people arrive in each city.

 

Example 1:

Input: [[10,20],[30,200],[400,50],[30,20]]
Output: 110
Explanation: 
The first person goes to city A for a cost of 10.
The second person goes to city A for a cost of 30.
The third person goes to city B for a cost of 50.
The fourth person goes to city B for a cost of 20.

The total minimum cost is 10 + 30 + 50 + 20 = 110 to have half the people interviewing in each city.
 

Note:

1 <= costs.length <= 100
It is guaranteed that costs.length is even.
1 <= costs[i][0], costs[i][1] <= 1000
'''

class Solution(object):
    def twoCitySchedCost(self, costs):
        """
        :type costs: List[List[int]]
        :rtype: int
        """
        result = 0
        costs = sorted(costs, key=lambda x : x[0] - x[1])
        for index in range(len(costs)):
            if index < len(costs)//2:
                result += costs[index][0]
            else:
                result += costs[index][1]
        return result
